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If Y is a continuous random variable and m is the median of the distribution, then m is such that P(Y ≤m) = P(Y ≥ m) = 1/2. If Y₁, Y₂,..., Y are independent, exponentially dis- tributed random variables with mean ß and median m, Example 6.17 implies that Y(n) max (Y₁, Y₂, ..., Yn) does not have an exponential distribution. Use the general form of Fy) (y) to show that P(Y(n) > m) = 1-(.5)".

If Y is a continuous random variable and m is the median of the distribution, then m is such that P(Y ≤m) = P(Y ≥ m) = 1/2. If Y₁, Y₂,..., Y are independent, exponentially dis- tributed random variables with mean ß and median m, Example 6.17 implies that Y(n) max (Y₁, Y₂, ..., Yn) does not have an exponential distribution. Use the general form of Fy) (y) to show that P(Y(n) > m) = 1-(.5)".

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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6.82 If Y is a continuous random variable and m is the median of the distribution, then m is such that P(Y ≤m) = P(Y ≥ m) = 1/2. If Y₁, Y₂, ..., Y, are independent, exponentially dis- tributed random variables with mean ß and median m, Example 6.17 implies that Y(n) = max (Y₁, Y₂, ..., Yn) does not have an exponential distribution. Use the general form of Fy) (y) to show that P(Y(n) > m) = 1- (.5)".
Transcribed Image Text:6.82 If Y is a continuous random variable and m is the median of the distribution, then m is such that P(Y ≤m) = P(Y ≥ m) = 1/2. If Y₁, Y₂, ..., Y, are independent, exponentially dis- tributed random variables with mean ß and median m, Example 6.17 implies that Y(n) = max (Y₁, Y₂, ..., Yn) does not have an exponential distribution. Use the general form of Fy) (y) to show that P(Y(n) > m) = 1- (.5)".
Expert Solution
Step 1

It is given that Y1, Y2, …,Yn are independent random variables that follow an exponential distribution with mean β and median m.

Thus the probability density function (PDF) of Y is as follows:

fYy=1βe-1βy, y≥0

As it is given that Y has median m, thus

FYm=PY≤m=12=0.5 

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